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How to arrive at an equation of a Roman surface from three points (a triangle)
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I have three points (for example, A( 1, 1, 0 ) B( 2, 1, 0 ) C( 1.5, 0, 0 ), origin at O( 0, 0, 1000 ) ), with which I am creating what I believe is a Roman surface ($x^2y^2 + y^2z^2 + z^2x^2 - r^2xyz = 0$). This surface has some useful properties in that it intersects itself in three lines that then intersect at its triple point at the origin of its radius. For my purposes I am imagining using a very large radius, because the surface will appear flat for points that are close together.
I believe that it is possible to describe a triangle with the pinch points of the intersecting surface as drawn in the illustration. Please excuse the crude lines, I was trying to describe the curvature of the surface as it obviously overlaps a lot ><.
Illustration of points on the surface
Additional, incorrect illustration
If I can derive an equation of a Roman surface that pinches at the three points, then it is trivial to prove if other points are within the surface. I'm not sure how to solve the equation due to its relative complexity. Intuitively, it seems something like a 3D analog of the Steiner circumellipse, which connects three points along a single "surface".
If I am incorrect in my assumptions about the Roman surface, is there another topology that might be useful here? The goal is to describe a surface that includes all three edges of the triangle. Thank you for your help, I'm quite out of my league.
general-topology algebraic-geometry algebraic-topology geometric-topology non-orientable-surfaces
asked Jul 24 at 23:32
Gavin
112
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up vote
2
down vote
favorite
I have three points (for example, A( 1, 1, 0 ) B( 2, 1, 0 ) C( 1.5, 0, 0 ), origin at O( 0, 0, 1000 ) ), with which I am creating what I believe is a Roman surface ($x^2y^2 + y^2z^2 + z^2x^2 - r^2xyz = 0$). This surface has some useful properties in that it intersects itself in three lines that then intersect at its triple point at the origin of its radius. For my purposes I am imagining using a very large radius, because the surface will appear flat for points that are close together.
I believe that it is possible to describe a triangle with the pinch points of the intersecting surface as drawn in the illustration. Please excuse the crude lines, I was trying to describe the curvature of the surface as it obviously overlaps a lot ><.
Illustration of points on the surface
Additional, incorrect illustration
If I can derive an equation of a Roman surface that pinches at the three points, then it is trivial to prove if other points are within the surface. I'm not sure how to solve the equation due to its relative complexity. Intuitively, it seems something like a 3D analog of the Steiner circumellipse, which connects three points along a single "surface".
If I am incorrect in my assumptions about the Roman surface, is there another topology that might be useful here? The goal is to describe a surface that includes all three edges of the triangle. Thank you for your help, I'm quite out of my league.
general-topology algebraic-geometry algebraic-topology geometric-topology non-orientable-surfaces
asked Jul 24 at 23:32
Gavin
112
For those who'd never heard of "Roman surface" before, and figured this was a typo for "Riemann surface"... as I did... see the Wiki page, for example... :)
– paul garrett
Jul 24 at 23:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have three points (for example, A( 1, 1, 0 ) B( 2, 1, 0 ) C( 1.5, 0, 0 ), origin at O( 0, 0, 1000 ) ), with which I am creating what I believe is a Roman surface ($x^2y^2 + y^2z^2 + z^2x^2 - r^2xyz = 0$). This surface has some useful properties in that it intersects itself in three lines that then intersect at its triple point at the origin of its radius. For my purposes I am imagining using a very large radius, because the surface will appear flat for points that are close together.
I believe that it is possible to describe a triangle with the pinch points of the intersecting surface as drawn in the illustration. Please excuse the crude lines, I was trying to describe the curvature of the surface as it obviously overlaps a lot ><.
Illustration of points on the surface
Additional, incorrect illustration
If I can derive an equation of a Roman surface that pinches at the three points, then it is trivial to prove if other points are within the surface. I'm not sure how to solve the equation due to its relative complexity. Intuitively, it seems something like a 3D analog of the Steiner circumellipse, which connects three points along a single "surface".
If I am incorrect in my assumptions about the Roman surface, is there another topology that might be useful here? The goal is to describe a surface that includes all three edges of the triangle. Thank you for your help, I'm quite out of my league.
general-topology algebraic-geometry algebraic-topology geometric-topology non-orientable-surfaces
asked Jul 24 at 23:32
Gavin
112
I have three points (for example, A( 1, 1, 0 ) B( 2, 1, 0 ) C( 1.5, 0, 0 ), origin at O( 0, 0, 1000 ) ), with which I am creating what I believe is a Roman surface ($x^2y^2 + y^2z^2 + z^2x^2 - r^2xyz = 0$). This surface has some useful properties in that it intersects itself in three lines that then intersect at its triple point at the origin of its radius. For my purposes I am imagining using a very large radius, because the surface will appear flat for points that are close together.
I believe that it is possible to describe a triangle with the pinch points of the intersecting surface as drawn in the illustration. Please excuse the crude lines, I was trying to describe the curvature of the surface as it obviously overlaps a lot ><.
Illustration of points on the surface
Additional, incorrect illustration
If I can derive an equation of a Roman surface that pinches at the three points, then it is trivial to prove if other points are within the surface. I'm not sure how to solve the equation due to its relative complexity. Intuitively, it seems something like a 3D analog of the Steiner circumellipse, which connects three points along a single "surface".
If I am incorrect in my assumptions about the Roman surface, is there another topology that might be useful here? The goal is to describe a surface that includes all three edges of the triangle. Thank you for your help, I'm quite out of my league.
general-topology algebraic-geometry algebraic-topology geometric-topology non-orientable-surfaces
asked Jul 24 at 23:32
Gavin
112
edited Jul 25 at 3:16
asked Jul 24 at 23:32
Gavin
112
asked Jul 24 at 23:32
Gavin
112
asked Jul 24 at 23:32
Gavin
112
112
For those who'd never heard of "Roman surface" before, and figured this was a typo for "Riemann surface"... as I did... see the Wiki page, for example... :)
– paul garrett
Jul 24 at 23:44
add a comment |Â
For those who'd never heard of "Roman surface" before, and figured this was a typo for "Riemann surface"... as I did... see the Wiki page, for example... :)
– paul garrett
Jul 24 at 23:44
For those who'd never heard of "Roman surface" before, and figured this was a typo for "Riemann surface"... as I did... see the Wiki page, for example... :)
– paul garrett
Jul 24 at 23:44
For those who'd never heard of "Roman surface" before, and figured this was a typo for "Riemann surface"... as I did... see the Wiki page, for example... :)
– paul garrett
Jul 24 at 23:44
add a comment |Â
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For those who'd never heard of "Roman surface" before, and figured this was a typo for "Riemann surface"... as I did... see the Wiki page, for example... :)
– paul garrett
Jul 24 at 23:44